Physics 2101 Section 3 May 7th: Chap. 20

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3) For liquids/solids - heat transfer: .... For a Carnot engine, from the 1st Law of thermo says that the net heat ... Otto cycle: 4-stroke internal combustion engine.
Physics 2101 Section 3 May 7th: Chap. 20 Announcements: • Final Exam: May 11th (Tuesday), 7:30 AM at HoweHoweRussell 130 • Make up Final: May 15th (Saturday) 7:30 AM at Nicholson 119 Chapter 20— 20—Irreversible processes

• Final Exam for those who need extended time will be at Nicholson 109 • Review: R i Saturday d 2PM at Nicholson 130

Class Website: http://www.phys.lsu.edu/classes/spring2010/phys2101--3/ http://www.phys.lsu.edu/classes/spring2010/phys2101 http://www.phys.lsu.edu/~jzhang/teaching.html

History of grades in 2101 for last 5 years:

The rough guide lines are  A:  0 positive TH ⎝ Va ⎠ ΔE int = 0



2) b → c: adiabatic expansion Q=0



ΔS b→c = 0

3) c → d: Isothermal compression

⎛V ⎞ QL = W = nRTL ln⎜ d ⎟ < 0 ⎝ Vc ⎠ ⎛ ⎞ Q V ΔS L = L = nR ln⎜ d ⎟ < 0 negative TL ⎝ Vc ⎠

ΔE int = 0

0 = ΔS cycle = ΔS H + ΔS L

4) d → a: adiabatic compression

Q Q 0= H − L TH TL

ε carnot = 1 −



Q=0

TL TH



ΔS d →a = 0

All Carnot engines operating between two constant  temperatures TH & TL have the same efficiency. An irreversible engine is less efficient  

Problem #1

W QH − QL QL ε= = =1− QH QH QH

ε carnot = 1 −

TL TH

Imagine a Carnot engine  that operates between the temperatures of TH = 850 K and TL = 300 K.   The engine performs 1200 J of work each cycle, which takes 0.25 s. a)

What is the efficiency of this engine?

ε =1−

The efficiency, ε, of a Carnot engine is ONLY determined by the ratio TL/TH: b)

What is the average power of this engine?

(

TL 300 K =1− ≅ 65% TH 850 K

)

Power is found from Work done per cycle/time per cycle: Pave = W Δt = 1200 J 0.25 s = 4800 W Power is found from Work done per cycle/time per cycle: c) How much energy |QH| is extracted as heat from the high temperature reservoir every  cycle? For any engine, the efficiency, ε, is defined as the work the engine does per cycle  y g y g p y divided by the energy it absorbs as heat per cycle: W W 1200 J ε= → QH = = ≅ 1855 J QH ε 65% d) How much energy |QL| is delivered as heat to the low temperature reservoir every cycle? F For a Carnot engine, from the 1 C t i f th 1st Law of thermo says that the net heat transfer per  L f th th t th th tt f cycle is equal to the  net work done: W = Q H − QL → QL = Q H − W = 1855 J −1200 J = 655 J e)

What is the entropy change ΔS of the working substance for the energy transfer to it from  the high‐temperature and low‐temperature reservoir? Along the hot and cool isotherms, the entropy changes are:  Q Q −655 J 1855 J → ΔS tot = 0 ΔS L = L = = −2.18 J/K ΔS H = H = = 2.18 J/K TL 300 K TH 850 K

Otto cycle: 4‐stroke internal combustion engine

a→ b : adiabatic compression stroke c → d : adiabatic expansion stroke

b → c : isochoric expansion due to fuel ignition d → a : isochoric compression due to exhaust valve

What is the efficiency of the Otto cycle? Assume the compression ratio of 4:1  (V4=4V1), the gas mixture can be regarded as  an ideal gas monatomic gas: id l t i Theoretical : 40% Reality : 20%

Problem

ε=

W Q − QL Q = H =1− L QH QH QH

Problem 20 Problem 20‐29 29 One mole of a monatomic ideal gas is taken through the  One mole of a monatomic ideal gas is taken through the reversible cycle shown. Vc=8.00Vb , pb=10.0 atm and Vb=10‐3m3.) What’s  the energy added to the gas as heat? b) the energy leaving the gas as  heat, c) What’s the net work done by the gas?  d) What’s the efficiency of  the cycle? the cycle? At the state b: PbVb = RTb so that Tb = PV b b /R b ⇒ c: Tb Vbγ -1 = Tc Vcγ -1 (γ = c ⇒ a ( p a = pc ) :

Cp

V T 5 1 = ), so that Tc = Tb ( b )γ −1 = Tb ( )γ −1 = γ b−1 Cv 3 Vc 8 8

Va Ta V V T T = ; so Ta = ( a )Tc = ( b )Tc = c = γb 8 8 Vc Tc Vc Vc

Energy added to the gas as heat:

Energy gy leaving g from the ggas as heat:

3R Qab = nCV ΔTab = (Tb − Ta ) 2

Qca = nC p ΔTca =

Net work: Wnet = Wbc + Wca = R (Tb + Ta − 2Tc ) Wbc = − ( ΔEin )bc = R (Tb − Tc ) Wca = pa ΔVca = R ΔTac = R (Ta − Tc )

5R (Tc − Ta ) 2

Problem #1 Entropy change with phase transition Entropy change with phase transition What is the entropy change of a 12.0 g ice cube that melts completely in a bucket of water whose  temperature is just above the freezing point of water?

How do we find the change in entropy?

ΔS = S f − S i =

f

∫ i

dQ T

In this case, what is the heat transferred to the system??

Qphase transition = mL

mL Plugging this in, we can calculate the entropy change:   ΔS = phase change (this is a reversible isothermal process!!) T Thus: 

ΔS ice =

(12 g)(333 J/g) = 14.6 J/K ≥ 0 (273 K)

Problem #2 Solid: entropy change with temperature A 50 0 g block of copper whose temperature is 400 K is placed in an insulating box with a 100 g A 50.0 g block of copper whose temperature is 400 K is placed in an insulating box with a 100 g block of lead whose temperature is 200 K (cCu=386 J/kg/K & cPb=128 J/kg/K ) a) What is the equilibrium temperature of the two‐block system? Review question We know from Chapt. 19 that in an insulating box, the net heat flow is zero, thus:

∑Q = 0 = m Tf =

c (T f − Ti,Cu )+m Pb cPb (T f − Ti,Pb )

Cu Cu

mCu cCuTi,Cu + m Pb cPb Ti,Pb = 320 K mCu cCu + m Pb cPb

b) What b)    What is the change in internal energy of the two‐block system is the change in internal energy of the two block system between the initial state and the between the initial state and the equilibrium state?    Since the system (Cu + Pb) is thermally insulated and no work is done, the  change in the internal energy is zero! c) 

What is the change in the entropy of the two‐block system? f

dQ ΔS = S − S = ∫ Knowing                                               what is the heat transfer? rev f i i T

dQheat capacity = mc(dT ) ΔS =

f

∫ i

⎛T ⎞ mc(dT ) = mc ln⎜ f ⎟ T ⎝ Ti ⎠

⎛ 320 K ⎞ ⎛ 320 K ⎞ ΔS tot = ΔSCu + ΔS Pb = mCucCu ln⎜ ⎟ + m Pb cPb ln⎜ ⎟ ⎝ 400 K ⎠ ⎝ 200 K ⎠ = − 4.3 J/K + 6.0 J/K = 1.7 J/K (positive!!)

Problem #3 Gas: entropy change during process An ideal gas undergoes an isothermal expansion at 77°C from 1.3 l to 3.4 l.  Then entropy change  during this process is 22 J/K.  How many moles of gas are there? What is the change in entropy during an isothermal process?

ΔS isothermal =

f

∫ i

dQ Q Q = T T

ΔE int,isothermal = 0

What do we know from 1st Law of Thermo?

ΔE int,isothermal = 0 ⇒ Q = W =



⎛Vf ⎞ ⎟ ⎝ Vi ⎠

pdV ⇒ Q = nRT ln⎜

Putting together, we find that:

n=

ΔS isothermal 22 J/K = = 2.75 moles ⎛Vf ⎞ ⎛ 3.4 l ⎞ 8 31 J/mole / l ⋅ K ) ln l ⎜ ⎟ R ln⎜ ⎟ (8.31 ⎝ ⎠ 1.3 l ⎝ Vi ⎠

Problem #4 Gas: entropy change during process One mole of an ideal monatomic gas is taken through the  cycle shown.   a) How much work is done by the gas in going from state  a to state c along abc? ⎡4V ⎤ W = W + W = p dV Work done is: net ⎢∫ 0 ⎥ a →b b →c ⎣V ⎦isochoric 0

0

⎡4V ⎤ + ⎢ ∫ pdV ⎥ ⎣4V ⎦isobaric 0

0

Wnet = Wa →b + Wb →c = (4V0 − V0 )p0 + 0 = 3V0 p0

What are the changes in internal energy and entropy going: b) from b to c ? Internal energy is given by change in temperature: ΔE int,b →c = nRΔT = nR(T f − Ti ) 3 2

3 2

⎛ 2 p ⋅ 4V0 p0 ⋅ 4V0 ⎞ = 32 nR⎜ 0 − ⎟ = 6 p0V0 ⎝ nR nR ⎠

c) through one complete cycle ? For a complete cycle: ΔE int,cycle = ΔS rev,cycle = 0

The entropy is found from: 

ΔS b →c

dQ f [dE int + Wby ] =∫ =∫ T i T i ⎡ f [32 nRdT + 0]⎤ ⎛ Tf ⎞ 3 = 2 nR ln⎜ ⎟ =⎢∫ ⎥ T isochoric ⎝ Ti ⎠ exp ⎣i ⎦isochoric ansion f

⎛T ⎞ ⎛8p V ⎞ = 32 R ln⎜ f ⎟ = 32 R ln⎜ 0 0 ⎟ ⎝ Ti ⎠ ⎝ 4 p0V0 ⎠ = 32 R ln (2)

19-17: Container A holds an ideal gas at a pressure of 5.0 x 105Pa at a T= 300 K. It is connected to a thin tube and a closed valve to container b, b with four times the volume of A. Container B holds the same ideal gas at a pressure of 1.0 x 105 Pa and T=400K. The valve is opened to allow the pressures to equalize, but the temperature of each container is maintained. What then is the pressure in the two containers?

Use ideal gas Law

After valve is opened

VA ⎛ pA 4 pB ⎞ n = na + nB = ⎜ + = Const. ⎟ R ⎝ TA TB ⎠ p 'A =

Rn ' A TA = p 'B VA

Rn ' B TB p 'B = 4VA

or

⎛ 4T ⎞ V ⎛p 4p ⎞ n = n ' A + n ' B = n ' A ⎜ 1 + A ⎟ = na + nB = A ⎜ A + B ⎟ TB ⎠ R ⎝ TA TB ⎠ ⎝

⎛ 4TA ⎞ n 'B = ⎜ n 'A ⎟ ⎝ T ⎠ B

n 'A =

V (pA / TA + 4 pB / TB ) R 1 + 4TA / TB

Final Answer Final Answer 4TA pA + pB n ' A RTA TB p' = = 4TA VA 1+ TB